英文:
How do I retain the outer XML of sub-elements?
问题
我有一个包含一系列<order/>元素的XML文件。我想将这个单一的XML文件拆分成多个文件，每个文件包含一个单独的订单。
示例输入文件：
<!-- language: lang-xml -->
<orders>
<order order-id="123">
<line-item product-id="ABC">ABC</line-item>
</order>
<order order-id="456">
<line-item product-id="DEF">DEF</line-item>
</order>
</orders>
期望的输出：
Order-123.xml:
<order order-id="123">
<line-item product-id="ABC">ABC</line-item>
</order>
Order-456.xml:
<order order-id="456">
<line-item product-id="DEF">DEF</line-item>
</order>
在这个阶段，我不关心将每个订单详细信息解组成结构体；我只想要每个<order/>节点的精确副本。
我尝试了一些使用xml:",innerxml"的变体，像这样：
<!-- language: lang-golang -->
type OrdersRaw struct {
Orders []OrderRaw `xml:"order"`
}
type OrderRaw struct {
Order string `xml:",innerxml"`
}
上面的代码中的问题是，每个Order值包含内部XML（从<line-item/>开始），但不包含包装的<order/>标签。
理想情况下，如果有一个xml:",outerxml"标签，它将满足我的需求。
我试图避免像手动连接内部XML和手写XML（例如<order order-id="123">）这样的笨拙方法。
英文:
I have an XML file that contains a list of individual <order/> elements. I would like to split this single XML file up into multiple files, each of which contain a single order.
Example input file:
<!-- language: lang-xml -->
<orders>
<order order-id="123">
<line-item product-id="ABC">ABC</line-item>
</order>
<order order-id="456">
<line-item product-id="DEF">DEF</line-item>
</order>
</orders>
Desired outputs:
> Order-123.xml:
>
> <order order-id="123">
> <line-item product-id="ABC">ABC</line-item>
> </order>
>
> Order-456.xml:
>
> <order order-id="456">
> <line-item product-id="DEF">DEF</line-item>
> </order>
At this stage, I am not concerned with unmarshalling each of the order details into a struct; I merely want an exact copy of each <order/> node in its own file.
I have tried a few variations of this using xml:",innerxml", like this:
<!-- language: lang-golang -->
type OrdersRaw struct {
Orders []OrderRaw `xml:"order"`
}
type OrderRaw struct {
Order string `xml:",innerxml"`
}
The problem in the above code is that each Order value contains the inner XML (starting with <line-item/>) but does not contain the wrapping <order/> tag.
Ideally, if there were an xml:",outerxml" tag, it would serve my purpose.
I am trying to avoid doing hacky things like manually concatenating the inner XML with handwritten XML (e.g. <order order-id="123">).
答案1
得分: 2
在你的OrderRaw结构体上使用XMLName：
type OrderRaw struct {
XMLName xml.Name `xml:"order"`
Order string `xml:",innerxml"`
OrderID string `xml:"order-id,attr"`
}
Playground: https://play.golang.org/p/onK5FExqzD.
**编辑：**如果你想保存所有属性，你需要使用MarshalerXML和UnmarshalerXML接口：
func (o *OrderRaw) UnmarshalXML(d *xml.Decoder, start xml.StartElement) error {
o.attrs = map[xml.Name]string{}
for _, a := range start.Attr {
o.attrs[a.Name] = a.Value
}
type order OrderRaw
return d.DecodeElement((*order)(o), &start)
}
func (o OrderRaw) MarshalXML(e *xml.Encoder, start xml.StartElement) error {
for name, attr := range o.attrs {
start.Attr = append(start.Attr, xml.Attr{Name: name, Value: attr})
}
start.Name = xml.Name{Local: "order"}
type order OrderRaw
return e.EncodeElement(order(o), start)
}
Playground: https://play.golang.org/p/XhkwqJFyMd.
英文:
Use XMLName on your OrderRaw:
type OrderRaw struct {
XMLName xml.Name `xml:"order"`
Order string `xml:",innerxml"`
OrderID string `xml:"order-id,attr"`
}
Playground: https://play.golang.org/p/onK5FExqzD.
EDIT: If you want to save all attributes, you'll have to use MarshalerXML and UnmarshalerXML interfaces:
func (o *OrderRaw) UnmarshalXML(d *xml.Decoder, start xml.StartElement) error {
o.attrs = map[xml.Name]string{}
for _, a := range start.Attr {
o.attrs[a.Name] = a.Value
}
type order OrderRaw
return d.DecodeElement((*order)(o), &start)
}
func (o OrderRaw) MarshalXML(e *xml.Encoder, start xml.StartElement) error {
for name, attr := range o.attrs {
start.Attr = append(start.Attr, xml.Attr{Name: name, Value: attr})
}
start.Name = xml.Name{Local: "order"}
type order OrderRaw
return e.EncodeElement(order(o), start)
}
Playground: https://play.golang.org/p/XhkwqJFyMd.
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论